A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
Total number of outcomes = 6 × 6 = 36 (i)Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) Hence, total number of favourable cases = 11 P (5 will come up either time) P (5 will not come up either time) (ii)Total number of cases, when 5 can come at least once = 11 P (5 will come at least once)
- 39
as the die is rolled twice
6*6=total number of outcomes
=36
now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
=11 outcomes.
so p(E)=favourable outcome /total outcome
p(E)=11/36
p(not E)=1-11/36
=36-11/36=25/36
probability of 5 not coming = 25/36
favourable outcomes=11
total outcome=36
p(E)=11/36
thus probability of 5 coming once=25/36
6*6=total number of outcomes
=36
now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
=11 outcomes.
so p(E)=favourable outcome /total outcome
p(E)=11/36
p(not E)=1-11/36
=36-11/36=25/36
probability of 5 not coming = 25/36
favourable outcomes=11
total outcome=36
p(E)=11/36
thus probability of 5 coming once=25/36
- 16
as the die is rolled twice
6*6=total number of outcomes
=36
now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
=11 outcomes.
so p(E)=favourable outcome /total outcome
p(E)=11/36
p(not E)=1-11/36
=36-11/36=25/36
probability of 5 not coming = 25/36
favourable outcomes=11
total outcome=36
p(E)=11/36
thus probability of 5 coming once=25/36
?
6*6=total number of outcomes
=36
now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
=11 outcomes.
so p(E)=favourable outcome /total outcome
p(E)=11/36
p(not E)=1-11/36
=36-11/36=25/36
probability of 5 not coming = 25/36
favourable outcomes=11
total outcome=36
p(E)=11/36
thus probability of 5 coming once=25/36
?
- 2
Give possible ("scientific") solutions for the identified problems for example:
1. The advantages of keeping correct animal numbers.
2. An example of a proper rotational grazing system.
3. Ways in which degraded areas (over grazed,bare areas, eroded,bush encroached) can be treated ( or which action can be implemented) to improve the veld condition.
Compile and discuss a simple cattle production plan. Which date will you mate them,when will the calves be born, when will you wean the calves.
1. The advantages of keeping correct animal numbers.
2. An example of a proper rotational grazing system.
3. Ways in which degraded areas (over grazed,bare areas, eroded,bush encroached) can be treated ( or which action can be implemented) to improve the veld condition.
Compile and discuss a simple cattle production plan. Which date will you mate them,when will the calves be born, when will you wean the calves.
- 0